Respuesta :
Answer:
the tension in the rope between the boxes is equal to 88 N
Explanation:
given,
the force applied on one body F = 176 N
When two bodies are moving on horizontal plane at constant velocity then their kinetic friction (f k) is equal to applied force F
According to newton third law the resultant force acting on one body is equal to the resultant force acting on the another body.
T is the tension in the rope
[tex]T- f_k = - (T- f_k)[/tex]
T - F = - (T - F)
T - 176 = - (T - 0)
2 T = 176
T = 176/2 = 88 N
so, the tension in the rope between the boxes is equal to 88 N
Answer:
88 N
Explanation:
Since the two boxes are identical, they have the same normal force and kinetic friction constant.
Also, since you know they are moving at a constant velocity, the force of 176 N being applied to one of the boxes must equal the sum of the force of friction from both boxes, that is:
176 = 2 * (F_n * u_k) where F_n is normal force and u_k is coefficient of kinetic friction.
Since the rope between the boxes only has to exactly cancel out 1 box worth of kinetic friction, you get:
F_t = (F_n * u_k) where F_t equals the tension in the rope.
Substitution F_t in the first equation leads to:
176 = 2 * (F_t)
F_t = 88
So the tension in the rope is exactly half of the 176 N force being applied, so it is 88 N.
